Question

求不定积分1/1+根号下x dx

Answer 1

令√x=t，则x=t²，dx=2tdt
故原式=2∫t/(1+t) dt
=2∫(t+1-1)/(t+1) dt
=2∫[1- 1/(t+1)]dt
=2t-2ln(t+1)+C
=2√x-2ln(√x+1)+C

Answer 2

let
x= tanu
dx=(secu)^2 du
∫√(1+x^2)/x dx
=∫ [secu/tanu] [(secu)^2 du]
=∫ [(secu)^3/tanu ] du
=∫ du/[ sinu. (cosu)^2 ]
=∫ cscu dtanu
= cscu.tanu + ∫ tanu. cscu.cotu du
= (1/cosu) + ∫ cscu du
= (1/cosu) + ln|cscu- cotu| + C
= √(1+x^2) + ln|√(1+x^2)/x -1/x | + C

Answer 3

Answer 4

解：令　√x-1=t 则x=(t+1)² dx=2(t+1)dt
∫x/(√x-1)dx=∫(t+1)²/t*2(t+1)dt
=2∫(t³+3t+3+1/t)dt
=2t³/3+3t² +6t+2lnt+C

Answer 5

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